Question 526746
# 1



{{{sqrt(n+12)-sqrt(n)>2}}}



{{{sqrt(n+12)>sqrt(n)+2}}}



{{{(sqrt(n+12))^2>(sqrt(n)+2)^2}}}



{{{n+12>(sqrt(n)+2)^2}}}



{{{n+12>n+4*sqrt(n)+4}}}



{{{n+12-n-4>4*sqrt(n)}}}



{{{8>4*sqrt(n)}}}



{{{4*sqrt(n)<8}}}



{{{sqrt(n)<8/4}}}



{{{sqrt(n)<2}}}



{{{(sqrt(n))^2<(2)^2}}}



{{{n<4}}}



Since we have {{{sqrt(n)}}}, this means that {{{n>=0}}} (to avoid taking the square root of a negative number)



Since {{{n<4}}} as well, we can combine these two inequalities to get the compound inequality:


{{{0<=n<4}}}


which is our answer.




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# 2


{{{3x+5 = x*sqrt(3)}}}



{{{5 = x*sqrt(3)-3x}}}



{{{5 = x*(sqrt(3)-3)}}}



{{{5/(sqrt(3)-3) = x}}}



{{{x = 5/(sqrt(3)-3)}}}



{{{x = (5(sqrt(3)+3))/((sqrt(3)-3)(sqrt(3)+3))}}}



{{{x = (5sqrt(3)+15)/(3-9)}}}



{{{x = (5sqrt(3)+15)/(-6)}}}



{{{x = -(5sqrt(3)+15)/(6)}}}


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# 3



{{{sqrt(h+3)+sqrt(h-1)=5}}}



{{{sqrt(h+3)=5-sqrt(h-1)}}}



{{{(sqrt(h+3))^2=(5-sqrt(h-1))^2}}}



{{{h+3=(5-sqrt(h-1))^2}}}



{{{h+3=25-10sqrt(h-1)+h-1}}}



{{{h+3=24-10sqrt(h-1)+h}}}



{{{h+3-24-h=-10sqrt(h-1)}}}



{{{-21=-10sqrt(h-1)}}}



{{{(-21)^2=(-10sqrt(h-1))^2}}}



{{{441=100(h-1)}}}



{{{441=100h-100}}}



{{{441+100=100h}}}



{{{541=100h}}}



{{{541/100=h}}}



{{{h=541/100}}}



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